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Chapter 5: Problem 27

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved \(30 \mathrm{~mm}\) horizontally.

Short Answer

Expert verified
The electron is deflected approximately 1.54 mm vertically.

Step by step solution

01

Determine Horizontal Travel Time

The horizontal distance traveled by the electron is given as 30 mm or 0.03 m. The horizontal speed of the electron is constant and equal to \(1.2 \times 10^{7} \ \text{m/s}\). Therefore, the time \( t \) it takes to cover this distance can be found using the formula \( t = \frac{d}{v}\), where \( d = 0.03 \ \text{m} \) and \( v = 1.2 \times 10^{7} \ \text{m/s}\).\[ t = \frac{0.03}{1.2 \times 10^{7}} \approx 2.5 \times 10^{-9} \ \text{s} \]
02

Calculate Vertical Acceleration

The vertical force acting on the electron is \(4.5 \times 10^{-16} \ \text{N}\). To find the vertical acceleration \(a\), use Newton's second law \( F = ma \), where \( m = 9.11 \times 10^{-31} \ \text{kg} \) is the mass of the electron.\[ 4.5 \times 10^{-16} = 9.11 \times 10^{-31} \cdot a \]Solving for \( a \), we get:\[ a = \frac{4.5 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 4.94 \times 10^{14} \ \text{m/s}^2 \]
03

Calculate Vertical Displacement

The vertical displacement (\(y\)) can be found using the equation for motion under constant acceleration: \( y = v_i t + \frac{1}{2} a t^2 \). Since the initial vertical speed \(v_i\) is 0, this simplifies to \( y = \frac{1}{2} a t^2 \).Substitute \( a = 4.94 \times 10^{14} \ \text{m/s}^2 \) and \( t = 2.5 \times 10^{-9} \ \text{s} \):\[ y = \frac{1}{2} \times 4.94 \times 10^{14} \times (2.5 \times 10^{-9})^2 \]Calculating this:\[ y = \frac{1}{2} \times 4.94 \times 10^{14} \times 6.25 \times 10^{-18} \approx 1.54 \times 10^{-3} \ \text{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law of Motion is a cornerstone of classical mechanics. It describes the relationship between the force applied to an object, the mass of that object, and the resulting acceleration. Mathematically, it is expressed as \[ F = ma \]where:
  • \( F \) is the force applied on the object (in Newtons),
  • \( m \) is the mass of the object (in kilograms),
  • \( a \) is the acceleration (in meters per second squared).
This equation implies that for a constant mass, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass.
In the context of an electron moving through a field, as described in the exercise, the vertical acceleration can be determined using the force exerted on the electron and its mass. This principle simplifies the understanding of how external forces influence electron motion and allows us to predict the resulting paths and speeds under given forces.
Electron Motion
When discussing electron motion, it is essential to consider its unique properties such as its small mass and negative charge. Electrons are subatomic particles found in atoms, and when they move, they can be influenced by electric and magnetic fields.Influences on Electrons:
  • Electric Forces: An electric field can exert force on electrons according to the equation \( F = qE \), where \( q \) is the charge of the electron and \( E \) is the magnitude of the electric field.
  • Magnetic Forces: When electrons move through magnetic fields, they experience a force perpendicular to both their velocity and the field, described by \( F = qvB \sin \theta \).
In this exercise, the electron's motion is affected by a constant vertical force, which indicates the presence of an external force such as an electric field causing a deflection from its path.
Understanding electron motion allows scientists and engineers to design and manipulate electron paths in devices such as cathode ray tubes, electron microscopes, and other electronic equipment.
Constant Acceleration
Constant acceleration describes a scenario where an object's acceleration does not change with time. This is a common simplifying assumption in physics that allows for easier computation of an object's motion.
When an object is under constant acceleration, its position can be predicted using the kinematic equations. For instance, if an object starts from rest, the relationship between distance \( y \), acceleration \( a \), and time \( t \) can be expressed as:\[y = \frac{1}{2} a t^2\]This formula is particularly useful in solving problems like the one at hand, where an electron moves horizontally while experiencing a vertical force that leads to constant vertical acceleration.
  • Initial Velocity: When the initial velocity in the direction of interest is zero, the equation simplifies to focusing solely on acceleration's effect.
  • Vertical Displacement: As in the exercise, this equation helps calculate the electron's eventual displacement due to vertical acceleration.
By applying the concept of constant acceleration, we can effectively predict and calculate how far an object will move vertically when influenced by a constant force.

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Most popular questions from this chapter

The tension at which a fishing line snaps is commonly called the line's "strength." What minimum strength is needed for a line that is to stop a salmon of weight \(85 \mathrm{~N}\) in \(11 \mathrm{~cm}\) if the fish is initially drifting at \(2.8 \mathrm{~m} / \mathrm{s}\) ? Assume a constant deceleration.

Using a rope that will snap if the tension in it exceeds \(387 \mathrm{~N}\), you need to lower a bundle of old roofing material weighing \(449 \mathrm{~N}\) from a point \(6.1 \mathrm{~m}\) above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13 \mathrm{~g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s},\) and its penetration depth was \(15 \mathrm{~mm} .\) If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

A firefighter who weighs \(712 \mathrm{~N}\) slides down a vertical pole with an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2},\) directed downward. What are the (a) magnitude and (b) direction (up or down) of the vertical force on the firefighter from the pole and the (c) magnitude and (d) direction of the vertical force on the pole from the firefighter?

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\) and \(\vec{F}_{2}=(-5 \mathrm{~N}) \hat{\mathrm{i}}+(8 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}} .\) What is the third force?
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